Let $A$ be $2 \times 2$ nonzero real matrix.which of the following is
true?
$(A)$ trace of $A^2$ is positive$(B)$ $A$ has non zero eigenvalue.
$(C)$ All entries of $A^2$ can't be negative.
$(D)$$A^2$ has at least one positive entry.
I tried to find examples to counter these statements.
I took $A$ as $\begin{pmatrix}1&2\\-3&2\end{pmatrix}$ and $A^2$ is $\begin{pmatrix}-5&6\\-9&-2\end{pmatrix}$
This cancels out option $A,C$
Now I am not sure how to figure out option $(B)$ and $(D)$
I tried to change numbers of $A$ to find the example that counters $(D)$ but it is time-consuming. Is there any fact that I am missing for $(B)$ and $(D)$? I think $A$ can have zero eigenvalues because in my experience I never saw any statement saying a matrix must have zero value to have an eigenvalue zero. So (D) is my last option to tick. What could be another way to solve this problem quickly?
Best Answer
Hint If $$A= \begin{bmatrix}0&1 \\ 0 &0 \end{bmatrix}$$ what is $A^2$?
Note that if $A= \begin{bmatrix}a&b \\ c &d\end{bmatrix}$ then $$A^2= \begin{bmatrix}a^2+bc& ab+bd \\ ac+cd &d^2+bc \end{bmatrix}= \begin{bmatrix}a^2+bc& b(a+d) \\ c(a+d) &d^2+bc \end{bmatrix}$$
If $bc \geq 0$ then the (1,1) entry of $A^2$ cannot be negative.
If $bc <0$ then $b,c$ have opposite signs. Then one of $b$ or $c$ has the same sign as $(a+d)$ making the (1,2) or (2,1) entry non-negative.